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  • A small object at rest, absorbs a light pulse of power 20 mW and duration 300 ns. Assuming speed of light as <img alt="3 \times 10^8 \mathrm{~m} / \mathrm{s}" src="https://entrancecorner.oncod

A small object at rest, absorbs a light pulse of power 20 mW and duration 300 ns. Assuming speed of light as 3 \times 10^8 \mathrm{~m} / \mathrm{s},the momentum of the object becomes equal to :

Option: 1

3 \times 10^{-17} \mathrm{~kg} \mathrm{~m} / \mathrm{s}


Option: 2

2 \times 10^{-17} \mathrm{~kg} \mathrm{~m} / \mathrm{s}


Option: 3

1 \times 10^{-17} \mathrm{~kg} \mathrm{~m} / \mathrm{s}


Option: 4

0.5 \times 10^{-17} \mathrm{~kg} \mathrm{~m} / \mathrm{s}


Answers (1)

best_answer

Method I-

Power =20 \mathrm{mw}

\mathrm{t}=300 \mathrm{nsec}

\text { energy absorbed }=300 \times 10^{-9} \times 20 \times 10^{-3}=6 \times 10^3 \times 10^{-12}=6 \times 10^{-9} \mathrm{~J}

\begin{aligned} & \text { Momentum }=\frac{\text { Energy }}{\mathrm{C}} \\ & =\frac{\text { Power } \times \text { time }}{\mathrm{C}} \\ & =\frac{\left(20 \times 10^{-3} \mathrm{w}\right)\left(300 \times 10^{-9} \mathrm{~s}\right)}{3 \times 10^8 \mathrm{~m} / \mathrm{s}} \\ & =2 \times 10^{-17} \mathrm{~kg}-\mathrm{m} / \mathrm{s} \end{aligned}

Method II-

                                                    

 

\begin{aligned} & \text { Pressure }=\frac{\text { Intensity }}{\mathrm{C}}=\frac{\text { Power }}{\text { Area } \times \mathrm{C}} \\ & \text { Pressure } \times \text { Area }=\frac{\text { Power }}{\mathrm{C}} \\ & \text { Force }=\frac{\text { Power }}{\mathrm{C}}=\frac{20 \times 10^{-3}}{3 \times 10^8} \\ & \mathrm{~F}=\frac{20}{3} \times 10^{-11} \mathrm{~N} \\ & \mathrm{~F} \Delta \mathrm{t}=\Delta \mathrm{P}(\text { momentum }) \\ & \frac{20}{3} \times 10^{-11} \times 300 \times 10^{-9}=\mathrm{P}_{\mathrm{f}}-\mathrm{P}_{\mathrm{i}} \\ \end{aligned}

\begin{aligned} & 20 \times 10^{-20} \times 100=\mathrm{P}_{\mathrm{f}} \\ & 2 \times 10^{-17}=\mathrm{P}_{\mathrm{f}} \end{aligned}

Posted by

manish painkra

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